A319045 Length of longest run of consecutive odd numbers having exactly n divisors.

1, 3, 1, 8, 1, 8, 1, 17, 1

Offset: 1

Author

Jon E. Schoenfield, Dec 22 2018

Keywords

• nonn
• more
• hard

Comments

a(n)=1 for n odd, since every number with an odd number of divisors is a square, and no two squares are consecutive odd numbers.

The start of the first run of exactly k consecutive odd numbers having exactly n divisors is A319046(n,k).

From David Wasserman, May 04 2019: (Start)

7 <= a(10) <= 8.

14 <= a(12) <= 59. Dickson's conjecture implies a(12) >= 39. Schinzel's Hypothesis H implies a(12) >= 41. (End)

Examples

			From _David Wasserman_, May 04 2019: (Start)
A run of 17 consecutive odd numbers with 8 divisors begins at 237805775327, so a(8) >= 17; a run of 18 or more consecutive odd numbers would include at least two that are multiples of 9, and every multiple of 9 having 8 divisors is also a multiple of 27, but the two multiples of 9 cannot both be multiples of 27, so a(8) = 17.
A run of 5 consecutive odd numbers with 14 divisors begins at 10943266106145622193005970311, so a(14) >= 5. A run of 6 consecutive odd numbers with 14 divisors would include at least two that are multiples of 3, and these two would differ by 6. These must be 3^13, 3^6*p for p prime > 3, or 3*p^6 for p prime > 3. But 3*p^6 = 3 (mod 27), while 3^13 and 3^6*p = 0 (mod 27), so no two of these can differ by 6. Therefore no such run exists, and a(14) = 5. (End)
		

Crossrefs

Cf. A119479 (analog for consecutive integers), A319046.

Extensions

a(6)-a(9) from David Wasserman, Feb 07 2019